There are ‘n’ points in a plane out of which ‘m points are collinear. How many different straight lines can form?
Examples:
Input : n = 3, m = 3 Output : 1 We can form only 1 distinct straight line using 3 collinear points Input : n = 10, m = 4 Output : 40
Number of distinct Straight lines = nC2 – mC2 + 1
How does this formula work?
Consider the second example above. There are 10 points, out of which 4 collinear. A straight line will be formed by any two of these ten points. Thus forming a straight line amounts to selecting any two of the 10 points. Two points can be selected out of the 10 points in nC2 ways.
Number of straight line formed by 10 points when no 2 of them are co-linear = 10C2…..…(i)
Similarly, the number of straight lines formed by 4 points when no 2 of them are co-linear = 4C2….(ii)
Since straight lines formed by these 4 points are same, straight lines formed by them will reduce to only one.
Required number of straight lines formed = 10C2– 4C2 + 1 = 45 – 6 + 1 = 40
Implementation of the approach is given as:
C++
// CPP program to count number of straight lines// with n total points, out of which m are// collinear.#include <bits/stdc++.h>using namespace std; // Returns value of binomial coefficient// Code taken from https://goo.gl/vhy4jpint nCk(int n, int k){ int C[k+1]; memset(C, 0, sizeof(C)); C[0] = 1; // nC0 is 1 for (int i = 1; i <= n; i++) { // Compute next row of pascal triangle // using the previous row for (int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j-1]; } return C[k];} /* function to calculate number of straight lines can be formed */int count_Straightlines(int n,int m){ return (nCk(n, 2) - nCk(m, 2)+1);} /* driver function*/int main(){ int n = 4, m = 3 ; cout << count_Straightlines(n, m); return 0;} |
Java
// Java program to count number of straight lines// with n total points, out of which m are// collinear.import java.util.*;import java.lang.*;public class GfG { // Returns value of binomial coefficient // Code taken from https://goo.gl/vhy4jp public static int nCk(int n, int k) { int[] C = new int[k + 1]; C[0] = 1; // nC0 is 1 for (int i = 1; i <= n; i++) { // Compute next row of pascal triangle // using the previous row for (int j = Math.min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1]; } return C[k]; } /* function to calculate number of straight lines can be formed */ public static int count_Straightlines(int n, int m) { return (nCk(n, 2) - nCk(m, 2) + 1); } // Driver function public static void main(String argc[]) { int n = 4, m = 3; System.out.println(count_Straightlines(n, m)); } // This code is contributed by Sagar Shukla} |
Python
# Python program to count number of straight lines# with n total points, out of which m are# collinear.# Returns value of binomial coefficient# Code taken from https://goo.gl/vhy4jpdef nCk(n, k): C = [0]* (k+1) C[0] = 1 # nC0 is 1 for i in range(1, n+1): # Compute next row of pascal triangle # using the previous row j = min(i, k) while(j>0): C[j] = C[j] + C[j-1] j = j - 1 return C[k]#function to calculate number of straight lines# can be formed def count_Straightlines(n, m): return (nCk(n, 2) - nCk(m, 2)+1)# Driven coden = 4m = 3print( count_Straightlines(n, m) );# This code is contributed by "rishabh_jain". |
C#
// C# program to count number of straight// lines with n total points, out of // which m are collinear.using System;public class GfG { // Returns value of binomial coefficient // Code taken from https://goo.gl/vhy4jp public static int nCk(int n, int k) { int[] C = new int[k + 1]; // nC0 is 1 C[0] = 1; for (int i = 1; i <= n; i++) { // Compute next row of pascal triangle // using the previous row for (int j = Math.Min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1]; } return C[k]; } // Function to calculate number of // straight lines can be formed public static int count_Straightlines(int n, int m) { return (nCk(n, 2) - nCk(m, 2) + 1); } // Driver Code public static void Main(String []args) { int n = 4, m = 3; Console.WriteLine(count_Straightlines(n, m)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to count number of straight lines// with n total points, out of which m are// collinear.// Returns value of binomial coefficientfunction nCk($n, $k){ $C = array_fill(0, $k + 1, NULL); $C[0] = 1; // nC0 is 1 for ($i = 1; $i <= $n; $i++) { // Compute next row of pascal triangle // using the previous row for ($j = min($i, $k); $j > 0; $j--) $C[$j] = $C[$j] + $C[$j-1]; } return $C[$k];}// function to calculate// number of straight lines// can be formed function count_Straightlines($n, $m){ return (nCk($n, 2) - nCk($m, 2) + 1);}// Driver Code$n = 4;$m = 3;echo(count_Straightlines($n, $m));// This code is contributed // by Prasad Kshirsagar?> |
Javascript
<script> // Javascript program to count number of straight // lines with n total points, out of // which m are collinear. // Returns value of binomial coefficient // Code taken from https://goo.gl/vhy4jp function nCk(n, k) { let C = new Array(k+1); C.fill(0); C[0] = 1; // nC0 is 1 for (let i = 1; i <= n; i++) { // Compute next row of pascal triangle // using the previous row for (let j = Math.min(i, k); j > 0; j--) C[j] = C[j] + C[j-1]; } return C[k]; } /* function to calculate number of straight lines can be formed */ function count_Straightlines(n,m) { return (nCk(n, 2) - nCk(m, 2)+1); } let n = 4, m = 3 ; document.write(count_Straightlines(n, m)); // This code is contributed by divyesh072019.</script> |
Output:
4
Time Complexity: O(max(n,m))
Auxiliary Space: O(1)
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
